Question

ΔABC is right angled at A and AD is perpendicular to BC. If AC = b and AB = c then find AD in terms of b and c.

CBSE class 10 - Triangles

Answer 1

$\tt \: If \triangle A B C \: \: is \: \: a \: \: right \: \: angle \Delta \\ \\\Rightarrow \quad So \quad B C^{2}=A B^{2}+A C^{2} \\\\\Rightarrow \quad B C^{2}=c^{2}+b^{2} \\\\\\ \tt Now, \\ \\ Area \: of \Delta A B C =\frac{1}{2} \times b \times h \\ \\ \tt =\frac{1}{2} \times(b) \times( c)$

Also,

$\begin{aligned} \text { Area of } \Delta & \text { ABC } \\ &=\frac{1}{2} \times b \times h \\ &=\frac{1}{2} \times B C \times A D \\ &=\frac{1}{2} \times\left(c^{2}+b^{2}\right) \times A D \end{aligned}$

So, from above two equation we have

Area of triangles are equal.

$\tt \quad \frac{1}{2} \times b \times c=\frac{1}{2} \times \left(c^{2}+b^{2}\right) \times A D \\ \\ \tt \Rightarrow \frac{1}{2} \times b \times c=\frac{1}{2} \times \left(c^{2}+b^{2}\right) \times a D \\ \\ \tt \Rightarrow b c=\left(c^{2}+b^{2}\right) \times A D \\ \\ \Rightarrow \dfrac{b c}{c^{2}+b^{2}}=A D$

Concepts Used:-

• Pythagoras Theorem

    $\tt \star Hypotenuse^2 = Base^2 + perpendicular^2$

• Area of triangle
• Equating two equations

Answer 2

Answer:

BC/√b²+c²

Step-by-step explanation:

triangle ABC is a right angled triangle,

so by using pythagoras property-

BC²=AC²+AB²

BC²=b²+c²

BC= √b²+c²

area of triangle ABC= 1/2 x base x height= 1/2 x b x c

also area of ∆ABC= 1/2 x √b²+c² x AD

so,1/2 x b x c= 1/2 x√b²+c²x AD

bc= √b²+c² x AD

AD = bc /√b²+c²